I hope this is clear. I am to find the value of a delta value > 0 such that |f(x) - L |< e when 0 < |x-a| < delta. I understand and can find the value of delta for limits such as lim -> -1 (2x-7) = -9, e = 0.1, delta = e/2 + .02/2 = 0.05. My problem is that I do not understand how to find delta > 0 for a limit x->2 x^3 = 8, e = .01. Can someone please show me how to write |f(x) - L|< e, -> |x^3-8|< .01. I tried using the difference of cubes |(x-2)(x^2 + 2x +4)|< .01. I do not know what to do from here, nor do I know if I set the absolute value less than .01 correctly. Can anybody help? The cube root and the square root functions are confusing me. Sorry, I couldn't insert the Greek letters in for epsilon nor delta.

Sep 26th 2009, 09:40 PM

Chris L T521

Quote:

Originally Posted by bosmith

I hope this is clear. I am to find the value of a delta value > 0 such that |f(x) - L |< e when 0 < |x-a| < delta. I understand and can find the value of delta for limits such as lim -> -1 (2x-7) = -9, e = 0.1, delta = e/2 + .02/2 = 0.05. My problem is that I do not understand how to find delta > 0 for a limit x->2 x^3 = 8, e = .01. Can someone please show me how to write |f(x) - L|< e, -> |x^3-8|< .01. I tried using the difference of cubes |(x-2)(x^2 + 2x +4)|< .01. I do not know what to do from here, nor do I know if I set the absolute value less than .01 correctly. Can anybody help? The cube root and the square root functions are confusing me. Sorry, I couldn't insert the Greek letters in for epsilon nor delta.